Area bounded by curve `y=sinx`, whit x-axis, when x varies from 0 to `(pi)/(2)` is :-

Options

A. 1 unit
B. 2 units
C. 3 unit
D. 0

Answer

To find the area bounded by the curve \( y = \sin x \) and the x-axis from \( x = 0 \) to \( x = \frac{\pi}{2} \), we will follow these steps:

### Step 1: Set up the integral
The area \( A \) under the curve \( y = \sin x \) from \( x = 0 \) to \( x = \frac{\pi}{2} \) is given by the integral:
\[ A = \int_{0}^{\frac{\pi}{2}} \sin x \, dx \]

### Step 2: Compute the integral
To compute the integral, we find the antiderivative of \( \sin x \):
\[ \int \sin x \, dx = -\cos x + C \]
Now we will evaluate the definite integral from \( 0 \) to \( \frac{\pi}{2} \):
\[ A = \left[-\cos x\right]_{0}^{\frac{\pi}{2}} \]

### Step 3: Evaluate the limits
Now we substitute the limits into the antiderivative:
\[ A = -\cos\left(\frac{\pi}{2}\right) - \left(-\cos(0)\right) \]
Calculating the cosine values:
\[ \cos\left(\frac{\pi}{2}\right) = 0 \quad \text{and} \quad \cos(0) = 1 \]
Thus, substituting these values gives:
\[ A = -0 - (-1) = 1 \]

### Step 4: Conclusion
The area bounded by the curve \( y = \sin x \) and the x-axis from \( x = 0 \) to \( x = \frac{\pi}{2} \) is:
\[ \boxed{1} \]

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